A Simple Derivation of Einstein's Mass-Energy Equation P.S.C. Bruskiewich Mathematical-physics, University of British Columbia, Vancouver, BC The Einstein's Mass-Energy Equation E - mc 2 can be derived in a number of ways. One simple derivation was shared with the author by Dr. F. A. Kaempffer and Dr. George Volkoff during the course of a pleasant conversation in Dr. Kaempffer's office in 1984. 1.0 A Mass absorbs Light Energy at Rest The Mass-Energy Equation E = mc 2 , a ubiquitous equation in modern physics, can be derived in a number of ways. One simple derivation comes from Einstein himself. Consider a mass M at rest on a frictionless table. Imagine two packets of light energy being absorbed by the mass in such a fashion that no net momentum is added to the mass M (refer to Fig. 1: The Mass at Rest Absorbing Two Packets of Light Energy). i/*E 2^1^ I M | ^5^1 ^ E Fig. 1: The Mass at Rest Absorbing Two Packets of Light Energy Each packet delivers momentum Ap to the mass M, however the two impulses are equal and opposite. The momentum delivered during a small time interval At comes from a beam of light of length Ax = cAt . From Maxwell's equations, the momentum delivered by each of the wave packets is Ap = FAt = (PA)At (E\ \ AAt = ( E \ AAx AAt J \cAAt j (-] AAt where A is the area of the light beam, P is the light pressure, E is the energy in the light beam and V is the volume of the light beam ( V = AAx = Ac At ). While at rest on the frictionless table, the mass M absorbs a total energy of E. Its net gain of momentum is zero. This is expected by the Conservation of Momentum. This simple derivation is interesting in its use of dimensional analysis, a technique not often used in th contemporary times, but was of second nature during the first half of the 20 century. 2.0 A Mass absorbs Light Energy While in Motion Now let a moving observer approach the table with speed v. By the Special Theory of Relativity the observer can consider himself at rest with the table moving towards him with velocity v. It appears now that the mass M is approaching him with velocity v and momentum Mv. The observer now sees the two light energy packets approaching the mass M at a slight angle (refer to Fig. 2: The Mass in Motion Absorbing Two Packets of Light Energy). 1/2 E ^9*f 1/2 E Fig. 2: The Mass in Motion Absorbing Two Packets of Light Energy In the observer's frame the packets of light appear to approach the mass M at a slope of v/c. In his frame each packet have momentum with a component in the direction of motion of * / ^ \ A v \c J By the conservation of momentum, the total apparent momentum before the absorption should be equivalent to the total apparent momentum after the absorption of the light, Mv + 2 \c J = (M + m)v where an apparent mass m is now added to the initial mass M. From this we find mv mc Acknowledgements: The author would like to posthumously thank Dr. F.A. Kaempffer and Dr. G.M. Volkoff for their discussion and encouragement in the writing of this short article. This paper was originally written in 1984 with their kind assistance. The manuscript remained lost in the author's papers for several decades, only recently rediscovered and published. Patrick Bruskiewich 2103